### Physics Problem: Swinging Child#### Problem Statement:A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42° from the vertical and releases her from rest.#### Questions:1. **Potential Energy:** - What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?2. **Speed at the Bottom:** - How fast will she be moving at the bottom of the swing?3. **Work Done by Tension:** - How much work does the tension in the ropes do as the child swings from the initial position to the bottom?#### Detailed Explanation:1. **Potential Energy:** - To determine the potential energy change, one must calculate the height difference between the initial position and the lowest point of the swing. The height \( h \) can be found using trigonometry where \( h = L - L \cos(\theta) \), with \( L = 2.20 \) meters and \( \theta = 42^\circ \). - The potential energy difference can be calculated using the formula for gravitational potential energy \( \Delta U = mgh \), where: - \( m \) is the mass of the child (25 kg), - \( g \) is the acceleration due to gravity (9.8 m/s²), - \( h \) is the height difference.2. **Speed at the Bottom:** - To find the speed at the bottom of the swing, use the conservation of energy principle where the potential energy at the highest point is converted to kinetic energy at the lowest point. - The kinetic energy \( \frac{1}{2}mv^2 = mgh \). Solving for \( v \), the velocity can be determined.3. **Work Done by Tension:** - Since the tension in the rope acts perpendicular to the direction of the swing's movement, no work is done by the tension force. This is because work \( W = F \cos(\theta) d \), and since \( \theta = 90^\circ \), \( \cos(90^\circ) = 0 \).Understanding these principles will help in analyzing the problem and finding the solution effectively.

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Description: ### Physics Problem: Swinging Child#### Problem Statement:A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42° from the vertical and releases her from rest.#### Questions:1. **Pote

Draw the free-body diagram of the system. Determine the height h.

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A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42° from the vertical and releases her from rest. (a) What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing? (b) How fast will she be moving at the bottom of the swing? (c) How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42° from the vertical and releases her from rest. (a) What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing? (b) How fast will she be moving at the bottom of the swing? (c) How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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### Physics Problem: Swinging Child

#### Problem Statement:
A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42° from the vertical and releases her from rest.

#### Questions:
1. **Potential Energy:**
- What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?

2. **Speed at the Bottom:**
- How fast will she be moving at the bottom of the swing?

3. **Work Done by Tension:**
- How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

#### Detailed Explanation:

1. **Potential Energy:**
- To determine the potential energy change, one must calculate the height difference between the initial position and the lowest point of the swing. The height \( h \) can be found using trigonometry where \( h = L - L \cos(\theta) \), with \( L = 2.20 \) meters and \( \theta = 42^\circ \).
- The potential energy difference can be calculated using the formula for gravitational potential energy \( \Delta U = mgh \), where:
- \( m \) is the mass of the child (25 kg),
- \( g \) is the acceleration due to gravity (9.8 m/s²),
- \( h \) is the height difference.

2. **Speed at the Bottom:**
- To find the speed at the bottom of the swing, use the conservation of energy principle where the potential energy at the highest point is converted to kinetic energy at the lowest point.
- The kinetic energy \( \frac{1}{2}mv^2 = mgh \). Solving for \( v \), the velocity can be determined.

3. **Work Done by Tension:**
- Since the tension in the rope acts perpendicular to the direction of the swing's movement, no work is done by the tension force. This is because work \( W = F \cos(\theta) d \), and since \( \theta = 90^\circ \), \( \cos(90^\circ) = 0 \).

Understanding these principles will help in analyzing the problem and finding the solution effectively.

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